I'm reading Woodhouse's book on classical mechanics and I got stuck on this problem:

A rigid body has angular velocity $\vec{\omega}$ and has one point $O$ fixed relative to a frame $\tilde{R}$. Show that if $\vec{\omega} \times \tilde{D}\vec{\omega} \neq 0$, then $O$ is the only point with zero acceleration relative to $\tilde{R}$.

Note. $\tilde{D}\vec{\omega}$ is the derivative with respect to time of $\vec{\omega}$ in $\tilde{R}$.

My approach. Suppose there exists $P$ in the rigid body with null acceleration and let $\vec{r}$ be the vector going from $O$ to $P$. Since the body is rigid, we have that the velocity $\vec{v}_P$ with respect to $\tilde{R}$ satisfies $$\vec{v}_P = \vec{v}_O + \vec{\omega} \times \vec{r} = \vec{\omega} \times \vec{r}$$ On differentiation with respect to time we get $$0 = \tilde{D}\vec{v}_P = \tilde{D}(\vec{\omega} \times \vec{r}) = (\tilde{D}\vec{\omega} \times \vec{r}) + (\vec{\omega} \times \tilde{D}\vec{r}) = (\tilde{D}\vec{\omega} \times \vec{r}) + (\vec{\omega} \times \vec{v}_P)$$ From this point on every manipulation I tried to prove that $\vec{r}$ must be the zero vector got me nowhere.

Does anyone know how to proceed?


Here is stream of conciousness attempt:

Take the scalar product of your expression with with $\omega$ to get $$ 0 = \omega\cdot (\dot \omega\times r)+ \omega\cdot (\omega\times v_P) $$ The last term is zero and $$ 0=\omega\cdot (\dot \omega\times r)= r\cdot (\omega\times \dot \omega) $$ So $r$ has no component parallel to $(\omega\times \dot \omega)$ and $r$ must lie in the plane defined by $\omega$ and $\dot \omega$.

Note added: actually we dont need the above. Start below:

From $$ 0 = (\dot \omega\times r)+ (\omega\times v) $$ and $$ (\omega\times v)=\omega \times(\omega\times r)= (\omega\cdot r)\omega- r |\omega|^2 $$ we see firtly that that $\dot \omega \times r$ is perpendicular to $r$ so taking the scalar product we get
$$ 0=(\omega\cdot r)^2-|r|^2||\omega|^2 $$ so $r$ is parallel to $\omega$ --- i.e $r=a\omega$ and this makes $v=(\omega\times r)=0$. Once we have $v=0$ then we have
$$ 0=(\dot \omega\times r)= a(\dot \omega\times \omega) $$ Thus $a$ is zero. Hence (because $\omega\ne 0$) $r$ is zero

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